Weak convergence of $\mathcal{U}([u_n,v_n])$ If $(u_n)_n$ and $(v_n)_n$ are two sequences of real numbers, such that $\forall n \in \mathbb{N},u_n<v_n,$ if $X_n$ is a random variable with density $f_{X_n}=\frac{1}{v_n-u_n}1_{[u_n,v_n]},$
and suppose that $(X_n)_n$ converges in distribution to X. Prove that $(u_n)_n$ and $(v_n)_n$ converges and find the distribution of $X.$
Can you tell me how to begin? We should use characteristic functions or distribution functions?
 A: First observe that 
$$
P\left(X_n>\frac{u_n+v_n}2\right)=1/2
$$
hence tightness of the sequence $(X_n)$ forces boundedness of $\left(\frac{u_n+v_n}2\right)_{n\geqslant 1}$.
Similarly, 
$$
P\left(X_n>\frac{u_n}4+\frac{3v_n}4\right)=1/4
$$
hence $\left(\frac{u_n}4+\frac{3v_n}4\right)_{n\geqslant 1}$ is also bounded. It follows that the sequences $(u_n)_n$ and $(v_n)_n$ are bounded. 
Take convergent subsequences to $u$ and $v$ respectively. If $u<v$, then $(X_{n_k})$ converges in distribution to a uniform law on $(u,v)$ and since $(X_n)$ is supposed to converge in distribution, the whole sequence converges in law to $\mathcal U(u,v)$ (this also implies that $u_n\to u$ and $v_n\to v$). If $u=v$, the limiting distribution is a Dirac mass at $u$.
There is also a connection with the convergence of types: let $a_n=1/(v_n-u_n)$ and $b_n=-u_na_n$. Then $a_nX_n+b_n$ has a uniform distribution on $[0,1]$. If $X$ is not degenerated, then the convergence of types theorem gives that $a_n\to 0$ and $b_n\to b$ hence $(u_n)$ and $(v_n)$ converge and the limiting distribution is uniform.
